# Brachistochrone with velocity - still a cycloid

## Homework Equations

Conservation of energy:
$$\frac{1}{2}mv^2-mgy=\frac{1}{2}mv_0^2$$

## The Attempt at a Solution

I know how to start the problem, but in the end have to solve the differential equation:
$$\frac{dy}{dx}=\sqrt{\frac{k^2-v_0^2-2gy}{v_0^2+2gy}}$$
which I can't solve. Any ideas and hints would be greatly appreciated!
You have to put at forward slash to close a latex instruction ie: /tex

Think of the initial velocity as the result of a vertical change in position of a height y0 along this path from a starting point of 0. Then you just have:

$$\frac{dy}{dx}=\sqrt{\frac{k^2-2gy0-2gy}{2gy0+2gy}}$$

which reduces to:

$$\frac{dy}{dx}=\sqrt{\frac{k^2-2g(y0+y)}{2g(y0+y)}}$$

$$\frac{dy}{dx}=\sqrt{\frac{k^2-2gy'}{2gy'}}$$

the solution to which is a cycloid.

AM